Class 10 Question Paper 2025 Maths Basic

ANNUAL EXAMINATION

Question Paper
2025
NCERT BASED SYLLABUS

FOR CBSE AND STATE BOARD
FOLLOWING NCERT

KVS QUESTION PAPERS

Series : GE1FH SET~1

- >
430/1/1
.
Roll No. Q.P. Code

- -
-
Candidates must write the Q.P. Code
on the title page of the answer-book.
> NOTE
(I) - (I) Please check that this question paper
27 contains 27 printed pages.

(II) - - (II) Q.P. Code given on the right hand side
- - of the question paper should be written
on the title page of the answer-book by
the candidate.
(III) - 38 (III) Please check that this question paper
contains 38 questions.

(IV) (IV) Please write down the Serial
, - Number of the question in the
answer-book at the given place
before attempting it.
(V) - 15 (V) 15 minute time has been allotted to read
- this question paper. The question paper
will be distributed at 10.15 a.m. From
10.15 10.15 10.15 a.m. to 10.30 a.m., the candidates
10.30 - will read the question paper only and
- will not write any answer on the
answer-book during this period.

#

MATHEMATICS (BASIC)
3 80
Time allowed : 3 hours Maximum Marks : 80

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(i) - 38
(ii) - , , ,
(iii) 1 18 (MCQ) 19 20
1
(iv) 21 25 - (VSA) 2
(v) 26 31 - (SA) 3
(vi) 32 35 - (LA) 5
(vii) 36 38 4
2
(viii) - , 2 , 2 ,
2 3
(ix) = ,

(x)

20 (MCQ) , 1 20 1 =20

1. a b (HCF) 1 ,
(LCM)
(A) a+b (B) a
(C) b (D) ab

2. 3+

(A) (B)

(C) (D)

3. x2 3x 2=0 (discriminant)

(A) 1 (B) 17

(C) (D)

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General Instructions :
Read the following instructions very carefully and strictly follow them :
(i) This question paper contains 38 questions. All questions are compulsory.
(ii) This question paper is divided into five Sections A, B, C, D and E.
(iii) In Section A, Questions no. 1 to 18 are Multiple Choice Questions (MCQs) and
questions number 19 and 20 are Assertion-Reason based questions of 1 mark
each.
(iv) In Section B, Questions no. 21 to 25 are Very Short Answer (VSA) type
questions, carrying 2 marks each.
(v) In Section C, Questions no. 26 to 31 are Short Answer (SA) type questions,
carrying 3 marks each.
(vi) In Section D, Questions no. 32 to 35 are Long Answer (LA) type questions
carrying 5 marks each.
(vii) In Section E, Questions no. 36 to 38 are case study based questions carrying
4 marks each. Internal choice is provided in 2 marks questions in each case
study.
(viii) There is no overall choice. However, an internal choice has been provided in
2 questions in Section B, 2 questions in Section C, 2 questions in Section D and
3 questions in Section E.
(ix) Draw neat diagrams wherever required. Take = wherever required, if not
stated.
(x) Use of calculator is not allowed.

SECTION A

This section has 20 Multiple Choice Questions (MCQs) carrying 1 mark each. 20 1=20

1. If the HCF of two positive integers a and b is 1, then their LCM is :
(A) a+b (B) a
(C) b (D) ab

2. The number 3 + is :
(A) a rational number (B) an irrational number
(C) an integer (D) a natural number

3. The discriminant of the quadratic equation x2 3x 2 = 0 is :
(A) 1 (B) 17
(C) (D)

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4. x+ = 3 (x 0) ax2 + bx + c = 0
a b+c
(A) 5 (B) 2
(C) 1 (D) 1

5. (3, 5)
(A) 8 (B) 2
(C) 2 (D) 8

6. - ,
(A) 1:2 (B) 2:1

(C) 1:1 (D) :2

7. - ?
(A) AAA (B) SSS
(C) SAS (D) RHS

8. , P - ?

(A) P = 60
(B) P = 80

(C) P = 40

(D) P

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4. The equation x + = 3 (x 0) is expressed as a quadratic equation in the

form of ax2 + bx + c = 0. The value of a b + c is :
(A) 5 (B) 2
(C) 1 (D) 1

5. For a point (3, 5), the value of (abscissa ordinate) is :
(A) 8 (B) 2
(C) 2 (D) 8

6. The mid-point of a line segment divides the line segment in the ratio :
(A) 1:2 (B) 2:1

(C) 1:1 (D) :2

7. Which of the following is not the criterion for similarity of triangles ?
(A) AAA (B) SSS
(C) SAS (D) RHS
8. From the figures given below, which of the following is true about the
measure of P ?

(A) P = 60

(B) P = 80
(C) P = 40

(D) The measure of P cannot be determined

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9. , O PA - OP = 10 cm , AP

(A) 10 cm

(B) 20 cm
(C) 5 cm

(D) 5 cm

10. - ?
(A) tan 45 = cot 45
(B) sin 90 = tan 45

(C) sin 30 = cos 30

(D) sin 45 = cos 45

11. :

(A) 1

(B) 1

(C) 0

(D) 1

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9. In the given figure, PA is a tangent to a circle with centre O. If
OP = 10 cm, then the length of AP is :

(A) 10 cm

(B) 20 cm

(C) 5 cm

(D) 5 cm

10. Which of the following statements is false ?

(A) tan 45 = cot 45

(B) sin 90 = tan 45

(C) sin 30 = cos 30

(D) sin 45 = cos 45

11. The value of is :

(A) more than 1

(B) 1

(C) 0

(D) 1

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12. , - ?

(A) x
(B) y
(C) z
(D) a

13.

(A) l
(B) l+a
(C) l + 2r
(D) l + 2r + a
14. (quadrant) ,
(A) 1:2
(B) 2:1
(C) 1:4
(D) 4:1

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12. In the given figure, which of the following angles represents the angle of
depression ?

(A) x
(B) y
(C) z
(D) a

13. The perimeter of the shaded region in the given figure is :

(A) l
(B) l+a
(C) l + 2r
(D) l + 2r + a

14. The ratio of the area of a quadrant of a circle to the area of the same
circle is :
(A) 1:2
(B) 2:1
(C) 1:4
(D) 4:1

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15. ?
(A)
(B)
(C)
(D)

16. -
- 10 25 25 40 40 55 55 70 70 85 85 100
2 3 7 6 6 6

(A) 40
(B) 55
(C) 47·5
(D) 62·5

17.

3000 4000 4000 5000 5000 6000 6000 7000
5 10 9 8

(A) 3000
(B) 4000
(C) 5000
(D) 6000

18. , - ?
(A) 1 6
(B) 7
(C) 7
(D) 7

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15. For which of the following solids is the lateral / curved surface area and
total surface area the same ?
(A) Cube
(B) Cuboid
(C) Hemisphere
(D) Sphere

16. The class mark of the median class of the following data is :

Class Interval 10 25 25 40 40 55 55 70 70 85 85 100
Frequency 2 3 7 6 6 6

(A) 40
(B) 55
(C) 47·5
(D) 62·5

17. The following distribution shows the number of runs scored by some
batsmen in test matches :

Runs Scored 3000 4000 4000 5000 5000 6000 6000 7000
Number of Batsmen 5 10 9 8
The lower limit of the modal class is :
(A) 3000
(B) 4000
(C) 5000
(D) 6000

18. In a random experiment of throwing a die, which of the following is a
sure event ?
(A) Getting a number between 1 and 6
(B) Getting an odd number < 7
(C) Getting an even number < 7
(D) Getting a natural number < 7

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19 20
(A) (R)
(A), (B), (C) (D)

(A) (A) (R) (R), (A)

(B) (A) (R) , (R), (A)

(C) (A) , (R)

(D) (A) , (R)

19. (A) : a b ,a b HCF, a b
LCM
(R) : HCF,

20. (A) : p , 4x + py + 8 = 0
2x + 2y + 2 = 0 ,4
(R) : a1x + b1y = c1 a2x + b2y = c2

, = = .

5 - (VSA) 2 5 2 =10

21. x y

+ = 1 x =3

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Questions number 19 and 20 are Assertion and Reason based questions. Two
statements are given, one labelled as Assertion (A) and the other is labelled as
Reason (R). Select the correct answer to these questions from the codes (A), (B),
(C) and (D) as given below.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the
correct explanation of Assertion (A).

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not
the correct explanation of Assertion (A).

(C) Assertion (A) is true, but Reason (R) is false.

(D) Assertion (A) is false, but Reason (R) is true.

19. Assertion (A) : For any two natural numbers a and b, the HCF of a and b
is a factor of the LCM of a and b.

Reason (R) : HCF of any two natural numbers divides both the
numbers.

20. Assertion (A) : The value of p for which the system of equations
4x + py + 8 = 0 and 2x + 2y + 2 = 0 is consistent is 4.

Reason (R) : The system of equations a 1x + b1y = c1 and a2x + b2y = c2
is consistent with infinitely many solutions, if
a1
= = .
a2

SECTION B
This section has 5 Very Short Answer (VSA) type questions carrying 2 marks
each. 5 2=10

21. Solve the following system of equations for x and y :
+ = 1 and x =3

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22. , PQ RS , POQ ~ SOR.

, OSR ~ OQP, ROQ = 125 ORS = 70 .
OSR OQP

23. 6 cm 10 cm ,
,
24. A B (0 A < 90 , 0 B < 90 ) ,
tan (A + B) = 1 tan (A B) =

tan 45 = 1.

25. 20 cm 60
= 3·14 = 1·73

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22. (a) In the given figure, if PQ RS, then prove that POQ ~ SOR.

OR

(b) In the given figure, OSR ~ OQP, ROQ = 125 and
ORS = 70 . Find the measures of OSR and OQP.

23. Two concentric circles are of radii 6 cm and 10 cm. Find the length of the
chord of the larger circle which touches the smaller circle.

24. (a) Find the values of A and B (0 A < 90 , 0 B < 90 ), if
tan (A + B) = 1 and tan (A B) = .

OR
(b) Prove that tan 45 = 1 geometrically.

25. A chord of a circle of diameter 20 cm subtends an angle of 60 at the
centre of the circle. Find the area of the corresponding minor segment of
the circle. (Use = 3·14 and = 1·73)

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6 - (SA) , 3 6 3 =18

26.

x

x, y, a b x

27. 0 9

28.
x + 3y = 6; 2x 3y = 12

x y x : y = 1 : 2.

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SECTION C
This section has 6 Short Answer (SA) type questions carrying 3 marks each. 6 3=18

26. (a) Prove that is an irrational number.
OR
(b) The factor tree of a number x is shown below :

Find the values of x, y, a and b. Hence, write the product of the
prime factors of the number x so obtained.

27. Find a quadratic polynomial whose sum and product of zeroes are 0 and
9, respectively. Also, find the zeroes of the polynomial so obtained.

28. (a) Solve the following system of equations graphically :
x + 3y = 6; 2x 3y = 12
OR
(b) x and y are complementary angles such that x : y = 1 : 2. Express
the given information as a system of linear equations in two
variables and hence solve it.

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29.

30.

=

31. 200 180

, ? 100
80 , 200

, ?

4 - (LA) , 5 4 5 =20

32. 180 ,
8

k 2x2 + kx + 3 = 0

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29. Prove that a rectangle circumscribing a circle is a square.

30. Prove that :

=

31. A lot consists of 200 pens of which 180 are good and the rest are
defective. A customer will buy a pen if it is not defective. The shopkeeper
draws a pen at random and gives it to the customer. What is the
probability that the customer will not buy it ? Another lot of 100 pens
containing 80 good pens is mixed with the previous lot of 200 pens. The
shopkeeper now draws one pen at random from the entire lot and gives it
to the customer. What is the probability that the customer will buy the
pen ?

SECTION D

This section has 4 Long Answer (LA) type questions carrying 5 marks each. 4 5=20

32. (a) The difference of the squares of two positive numbers is 180. The
square of the smaller number is 8 times the greater number. Find
the two numbers.

OR

(b) Find the value(s) of k for which the equation 2x2 + kx + 3 = 0 has
real and equal roots. Hence, find the roots of the equations so
obtained.

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33.

ABCD , AC BD - O

= ABCD

34.

5 cm ,

3·5 cm
, ?

35. 200

0 20 10
20 40 35
40 60 50
60 80 60
80 100 30
100 120 15

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33.
In a quadrilateral ABCD, diagonals AC and BD intersect each other at O
such that = as shown in the given figure. Prove that ABCD is a
trapezium.

34. (a) A toy is in the form of a cone surmounted on a hemisphere. The
cone and hemisphere have the same radii. The height of the conical
part of the toy is equal to the diameter of its base. If the radius of
the conical part is 5 cm, find the volume of the toy.
OR

(b) A cubical block is surmounted by a hemisphere of radius 3·5 cm.
What is the smallest possible length of the edge of the cube so that
the hemisphere can totally lie on the cube ? Find the total surface
area of the solid so formed.

35. The following data gives the information on the observed lifetime (in
hours) of 200 electrical components :

Lifetime Number of electrical
(in hours) components
0 20 10
20 40 35
40 60 50
60 80 60
80 100 30
100 120 15

Find the mean lifetime (in hours) of the electrical components.

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3 , 4 3 4 =12

1

36. 15 m
,

60

,
(i) , 1

(ii) , 1

(iii) ,
, 30

-
2

,
2

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SECTION E

This section has 3 case study based questions carrying 4 marks each. 3 4=12

Case Study 1
36. An injured bird was found on the roof of a building. The building is 15 m
high. A fireman was called to rescue the bird. The fireman used an
adjustable ladder to reach the roof. He placed the ladder in such a way
that the ladder makes an angle of 60 with the ground in order to reach
the roof.

Based on the above information, answer the following questions :
(i) Find the length of the ladder used by the fireman to reach the roof. 1
(ii) Find the distance of the point on the ground at which the ladder
was fixed from the bottom of the building. 1
(iii) In order to avoid skidding, the fireman placed the ladder in such a
way that the bottom of the ladder touches the base of the wall
which is opposite to the building, making an angle of 30 with the
ground.
(a) Draw a neat diagram to represent the above situation and
hence find the width of the road between the building and
the wall. 2
OR
(b) Find the length of the ladder used by the fireman in this
case. 2

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2

37. ,
, A B ,
50 cm, 100 cm, 150 cm, ....... A ,
1 10 , 2 20 , 3 30

,

(i) 13 ? 1

(ii) n 500 cm , n 1

(iii) 11 ? 2

- , 450 ? 2

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Case Study 2

37. In a garden, saplings of rose flowers were planted at equal intervals to
form a spiral pattern. The spiral is made up of successive semicircles,
with centres alternatively at A and B, starting with centre at A, of radii
50 cm, 100 cm, 150 cm, ....... as shown in the figure given below. Spiral 1
has 10 flowers, Spiral 2 has 20 flowers, Spiral 3 has 30 flowers and so on.

Based on the above information, answer the following questions :

(i) What is the radius of the 13th spiral ? 1

(ii) If the radius of the nth spiral is 500 cm, find the value of n. 1

(iii) (a) Find the total number of saplings till the 11th spiral. 2

OR

(b) Till which spiral, will there be a total of 450 saplings ? 2

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3

38. A(10, 20)
B(50, 50) , P Q,

AB , AP = PQ = QB.

,

(i) C 1

(ii) 1

(iii) P 2

A Q 2

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Case Study 3

38. In a society, there is a circular park having two gates. The gates are
placed at points A(10, 20) and B(50, 50), as shown in the figure below.
Two fountains are installed at points P and Q on AB such that
AP = PQ = QB.

Based on the above information, answer the following questions :

(i) Find the coordinates of the centre C. 1

(ii) Find the radius of the circular park. 1

(iii) (a) Find the coordinates of the point P. 2

OR

(b) Find the distance of the fountain at Q from gate A. 2

430/1/1 # 27| P a g e P.T.O.

Marking Scheme Strictly Confidential
(For Internal and Restricted use only) Secondary School Examination, 2025
SUBJECT NAME MATHEMATICS (BASIC) (Q.P. CODE 430/1/1)

General Instructions: -

1 You are aware that evaluation is the most important process in the actual and correct assessment of the
candidates. A small mistake in evaluation may lead to serious problems which may affect the future
of the candidates, education system and teaching profession. To avoid mistakes, it is requested that
before starting evaluation, you must read and understand the spot evaluation guidelines carefully.
2 “Evaluation policy is a confidential policy as it is related to the confidentiality of the examinations
conducted, evaluation done and several other aspects. It’s leakage to public in any manner could
lead to derailment of the examination system and affect the life and future of millions of
candidates. Sharing this policy/document to anyone, publishing in any magazine and printing in
News Paper/Website etc. may invite action under various rules of the Board and IPC.”
3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done
according to one’s own interpretation or any other consideration. Marking Scheme should be strictly
adhered to and religiously followed. However, while evaluating, answers which are based on
latest information or knowledge and/or are innovative, they may be assessed for their
correctness otherwise and due marks be awarded to them. In class-X, while evaluating two
competency-based questions, please try to understand given answer and even if reply is not
from marking scheme but correct competency is enumerated by the candidate, due marks
should be awarded.
4 The Marking scheme carries only suggested value points for the answers.
These are in the nature of Guidelines only and do not constitute the complete answer. The students can
have their own expression and if the expression is correct, the due marks should be awarded accordingly.
5 The Head-Examiner must go through the first five answer books evaluated by each evaluator on the
first day, to ensure that evaluation has been carried out as per the instructions given in the Marking
Scheme. If there is any variation, the same should be zero after deliberation and discussion. The
remaining answer books meant for evaluation shall be given only after ensuring that there is no
significant variation in the marking of individual evaluators.
6 Evaluators will mark (✓) wherever answer is correct. For wrong answer CROSS ‘X” be
marked. Evaluators will not put right (✓) while evaluating which gives an impression that answer is
correct and no marks are awarded. This is most common mistake which evaluators are
committing.
7 If a question has parts, please award marks on the right-hand side for each part. Marks awarded for
different parts of the question should then be totaled up and written in the left-hand margin and
encircled. This may be followed strictly.
8 If a question does not have any parts, marks must be awarded in the left-hand margin and encircled.
This may also be followed strictly.
9 If a student has attempted an extra question, answer of the question deserving more marks should be
retained and the other answer scored out with a note “Extra Question”.
10 No marks to be deducted for the cumulative effect of an error. It should be penalized only once.
11 A full scale of marks (example 0 to 80/70/60/50/40/30 marks as given in Question Paper) has to be
used. Please do not hesitate to award full marks if the answer deserves it.
12 Every examiner has to necessarily do evaluation work for full working hours i.e., 8 hours every day
and evaluate 20 answer books per day in main subjects and 25 answer books per day in other subjects
(Details are given in Spot Guidelines).This is in view of the reduced syllabus and number of questions
in question paper.
430/1/1 1 P.T.O.

13 Ensure that you do not make the following common types of errors committed by the Examiner in the
past:-
Leaving answer or part thereof unassessed in an answer book.
Giving more marks for an answer than assigned to it.
Wrong totaling of marks awarded on an answer.
Wrong transfer of marks from the inside pages of the answer book to the title page.
Wrong question wise totaling on the title page.
Wrong totaling of marks of the two columns on the title page.
Wrong grand total.
Marks in words and figures not tallying/not same.
Wrong transfer of marks from the answer book to online award list.
Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly and
clearly indicated. It should merely be a line. Same is with the X for incorrect answer.)
Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
14 While evaluating the answer books if the answer is found to be totally incorrect, it should be marked
as cross (X) and awarded zero (0) Marks.
15 Any unassessed portion, non-carrying over of marks to the title page, or totaling error detected by the
candidate shall damage the prestige of all the personnel engaged in the evaluation work as also of the
Board. Hence, in order to uphold the prestige of all concerned, it is again reiterated that the
instructions be followed meticulously and judiciously.
16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for
spot Evaluation” before starting the actual evaluation.
17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title page,
correctly totaled and written in figures and words.
18 The candidates are entitled to obtain photocopy of the Answer Book on request on payment of
the prescribed processing fee. All Examiners/Additional Head Examiners/Head Examiners are once
again reminded that they must ensure that evaluation is carried out strictly as per value points for each
answer as given in the Marking Scheme.

430/1/1 2

430/1/1
MARKING SCHEME
MATHEMATICS (BASIC)

Answer : (D) ab 1

Answer : (B) an irrational number 1

Answer : (B) 17 1

Answer : (A) 5 1

Answer : (D) 8 1

Answer : (C) 1 : 1 1

430/1/1 3 P.T.O.

Answer : None of the given options is correct. 1
Note – One mark to be given to all students who have attempted this question.

Answer : (C)  P = 40 1

Answer : (D) 5√3 cm 1

430/1/1 4

Answer : (C) sin 30 = cos 30 1

Answer : (D) – 1 1

Answer : (C) z 1

430/1/1 5 P.T.O.

Answer : (C) l + 2r 1

Answer : (C) 1 : 4 1

Answer : (D) Sphere 1

Answer : (D) 62·5 1

430/1/1 6

Answer : (B) 4000 1

Answer : (D) Getting a natural number < 7 1

Answer : (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct 1
explanation of Assertion (A).

430/1/1 7 P.T.O.

Answer : (D) Assertion (A) is false, but Reason (R) is true. 1

Solution : Solving the two equations to get x = 2, y = – 3 1+1

Solution : (a) As PQ ∥ RS
P=S
Alternate interior angles
Q=R 1
 POQ  SOR (by AA similarity criterion) 1
OR
(b)  OSR = 125 – 70 = 55 [ by exterior angle property] 1
As  OSR   OQP
 OSR =  OQP (Corresponding angles of similar triangles) ½
½
 OQP = 55

430/1/1 8

Solution:

Correct
figure
½

BC2 = OB2 – OC2
⇒ BC2 = 102 – 62 = 64
⇒ BC = 8 cm 1
AB = 8  2 = 16 cm ½

Solution: (a) A + B = 45 ½
A – B = 30 ½
Solving and getting A = 37·5 and B = 7·5 1
OR
(b) Consider an isosceles right  ABC ½
Using angle sum property A =  C = 45 ½
𝐴𝐵 𝑥
Clearly, tan 45 = = =1 1
𝐵𝐶 𝑥

Solution: Radius of circle = 10 cm = r
 OPQ is an equilateral triangle. ½
1 2 – √3  (10)2
Area of segment =  (3.14)  (10) 1
6 4
109
= sq. cm or 9.08 sq. cm
12 ½

430/1/1 9 P.T.O.

p
Solution: (a) Let √3 be a rational number such that √3 = (p and q are co-prime numbers, q  0) ½
q
√3q = p  3q2 = p2
3 divides p2  3 divides p as well
1
Let, p = 3m (for some integer m)
3q2 = 9m2  q2 = 3m2
3 divides q2  3 divides q as well 1
p and q have a common factor 3, which is a contradiction as p and q are co-prime.
 our assumption is wrong ½
Hence, √3 is an irrational number
OR
(b) b = 7 ½
a=3 ½
y = 420 ½
x = 840 ½
1
x = 840 = 23  3  5  7

Solution: Polynomial is x2 – 0 (x) + (– 9) = x2 – 9 1
For zeroes :
x2 – 9 = (x + 3) (x – 3) 1
Zeroes are – 3, 3 1

430/1/1 10

Solution: (a) Correct graph of each equation 1+1

Solution is x = 6, y = 0 or (6, 0) 1
OR
(b) x + y = 90 1
2x = y 1
Solving to get x = 30, y = 60 ½+½

Solution:

Correct
figure ½

As the length of tangents from an external point to a circle are equal
Thus,
AP = AS
BP = BQ
DR = DS
CR = CQ 1
Adding the above equations,
AB+CD=BC+AD
As AB =CD & BC = AD (opp. sides of rectangle)
 AB = AD 1
 ABCD is a square ½

𝑐𝑜𝑠2 𝐴 𝑠𝑖𝑛2 𝐴+𝑐𝑜𝑠2 𝐴
1+ 𝑐𝑜𝑠 2 𝐴
𝑠𝑖𝑛2 𝐴 𝑠𝑖𝑛2 𝐴 1
Solution: LHS = 𝑠𝑖𝑛2 𝐴
= 𝑐𝑜𝑠2 𝐴+𝑠𝑖𝑛2 𝐴
=
1+ 𝑠𝑖𝑛2 𝐴
𝑐𝑜𝑠2 𝐴 𝑐𝑜𝑠2 𝐴
𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴 2
𝑐𝑜𝑠 2 𝐴 𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴 2 𝑠𝑖𝑛𝐴 1½
= (𝑐𝑜𝑠𝐴−𝑠𝑖𝑛𝐴) = ( 𝑐𝑜𝑠𝐴−𝑠𝑖𝑛𝐴 )
𝑠𝑖𝑛2 𝐴
𝑐𝑜𝑠𝐴
1−𝑐𝑜𝑡𝐴 2
=( ) = RHS ½
1−𝑡𝑎𝑛𝐴
430/1/1 11 P.T.O.

20 1 1
Solution: P (customer will not buy the pen) = 200 = 10
After mixing the two lots
Total pens = 200 + 100 = 300
Number of good pens = 180 + 80 = 260 1
260 13
P (customer will buy the pen) = 300 or 15 1

Solution: (a) Let the smaller number be y and greater number be x.
A.T.Q.
x2 – y2 = 180 1
y2 = 8x 1
 x2 – 8x = 180
x2 – 8x – 180 = 0 1
(x – 18) (x + 10) = 0
x = 18, x = – 10 (rejected) 1
∴ The numbers are 18 and 12 1

OR
(b) For equal roots; b2 – 4ac = 0
k2 – 24 = 0 1
 k =  2√6 1
Equations are
2x2 + 2√6x + 3 = 0; 2x2 – 2√6x + 3 = 0
3 3 3 3
½+½
Roots are x = −√2, −√2; x = √2 , √2
1+1
430/1/1 12

Solution: Statement: If a line is drawn parallel to one side of a triangle to intersect the other two
sides in distinct points, the other two sides are divided in the same ratio. 1
𝐴𝑂 𝐶𝑂 Correct
Given : ABCD is a quadrilateral in which 𝐵𝑂 = 𝐷𝑂 figure,
given, to
To Prove : AB ∥ CD prove and
construction
Construction : Draw OE ∥ AB
1
Proof : In  DAB, OE ∥ AB
𝐷𝐸 𝐷𝑂
∴ 𝐴𝐸 = 𝐵𝑂 (by BPT) 1
𝐴𝑂 𝐶𝑂
Also 𝐵𝑂 = 𝐷𝑂 (given)
𝐷𝑂 𝐶𝑂
 𝐵𝑂 = 𝐴𝑂

𝐷𝐸 𝐶𝑂
 𝐴𝐸 = 𝐴𝑂 ½
In  ADC,
𝐷𝐸 𝐶𝑂
=
𝐴𝐸 𝐴𝑂
 OE ∥ CD (by converse of BPT) 1
As OE ∥ AB and OE ∥ CD
 AB ∥ CD ½
Hence, ABCD is a trapezium

Solution: (a) Radius = r = 5 cm
Height of cone = h = 10 cm
Volume of toy = volume of hemisphere + volume of cone
2
= r3+ 3r2h
1
3
2 22 1 22
= 3  7  5  5 5 + 3  7  5  5  10 2+2
5500 5500
= 21 + 21
430/1/1 13 P.T.O.

11000 1
= cu. cm or 523.81 cu. cm
21
OR
(b) Edge of cube = a = 3·5  2 = 7 cm 1
Total surface area of solid
= 6 a2 + 2r2 – r2
= 6 a2 + r2
22 1½ + 1½
= 6  7  7 + 7  3·5  3·5
665 1
= 2 sq. cm or 332·5 sq. cm

Solution:
𝑥 –𝑎
CI xi fi ui = 𝑖20 fiui
0 – 20 10 10 –2 – 20 Correct
20 – 40 30 35 –1 – 35 table
40 – 60 50 = a 50 0 0 2
60 – 80 70 60 1 60
80 – 100 90 30 2 60
100 – 120 110 15 3 45
Total 200 110
110
Mean = 50 + 200  20 2
Mean = 61 1
Thus the mean lifetime of the electrical components is 61 hours.

430/1/1 14

Solution: (i) Let the length of the ladder be ‘a’
15
= sin 60 ½
𝑎
30
a= or 10√3 ½
√3
30
Thus the length of the ladder is m or 10√3 m
√3

(ii) Let the distance of the point on the ground be ‘x’
15
= tan 60 ½
𝑥
15
x= or 5√3 ½
√3
15
Thus, the distance of the point on the ground is m or 5√3 m
√3

(iii) (a) Let the width of the road be y.

Correct
figure
1

430/1/1 15 P.T.O.

15
= tan 30
𝑦 ½
y = 15√3
Thus, the width of the road is 15√3 m. ½
OR
(b) Let the length of the ladder be l.

15
= sin 30
𝑙
l = 30 1
Thus, the length of the ladder is 30 m. 1

430/1/1 16

Solution: (i) a13 = 650 cm 1
(ii) an = 500
50 + (n – 1)50 = 500
n = 10 1
(iii) (a) a = 10, d = 10
11 1½
S11 = 2 [20 + 10  10]
= 660 ½
OR
(b) a = 10, d = 10
𝑛 1
450 = 2 [20 + (n – 1) 10]
½
n2 + n – 90 = 0
n=9 ½

10 + 50 20 + 50
Solution: (i) Co-ordinates of C are ( , ) = C(30, 35) 1
2 2
(ii) Radius = √(30 – 10)2 + (35 – 20)2 = 25 1
(iii) (a) P divides AB in the ratio 1 : 2, ½

430/1/1 17 P.T.O.

1 × 50 + 2 × 10 1 × 50 + 2 × 20 ½+½
co-ordinates of P are ( , )
3 3

70 ½
i.e. ( 3 , 30)
OR
(b) Distance AB = 2  25 = 50 ½
2 2
AQ = 3AB = 3 50 1
100
AQ = 3 ½

430/1/1 18

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